Let $$A = \left( {1,\,2} \right),\,B = \left( {3,\,4} \right)$$     and let $$C = \left( {x,\,y} \right)$$   be a point such that $$\left( {x - 1} \right)\left( {x - 3} \right) + \left( {y - 2} \right)\left( {y - 4} \right) = 0.$$        If ar $$\left( {\Delta ABC} \right) = 1$$   then maximum number of positions of $$C$$ in the $$x-y$$  plane is :

Solution :

$$\eqalign{ & \left( {x - 1} \right)\left( {x - 3} \right) + \left( {y - 2} \right)\left( {y - 4} \right) = 0 \cr & \Rightarrow AC \bot BC\,\,\,\,\,\, \Rightarrow \angle ACB = {90^ \circ } \cr & \therefore C{\text{ is on the circle whose diameter is }}AB \cr & AB = 2\sqrt 2 \cr & {\text{As ar }}\left( {\Delta ABC} \right) = 1,\,\frac{1}{2}.2\sqrt 2 .\left( {{\text{altitude}}} \right) = 1 \cr & \therefore {\text{altitude}} = \frac{1}{{\sqrt 2 }}{\text{ < radius}}{\text{.}} \cr & {\text{So, there are four possible positions of }}C. \cr & {\bf{Note:}} \cr & {\text{If ar}}\left( {\Delta ABC} \right) = 2,{\text{ altitude}} = {\text{radius }}\, \Rightarrow {\text{ two possible positions of }}C. \cr} $$